Question

In each of Problems 1 through 8 determine whether the given pair of functions is linearly independent or linearly dependent. \(f(\theta)=\cos 3 \theta\) \(\quad\) \(g(\theta)=4 \cos ^{3} \theta-3 \cos \theta\)

Short answer

Given the functions \(f(\theta) = \cos 3 \theta\) and \(g(\theta) = 4 \cos^3 \theta - 3 \cos \theta\), we applied the Wronskian test to determine whether they are linearly independent or dependent. After calculating and simplifying their Wronskian, we found that it is nonzero at \(\theta = \frac{\pi}{12}\). Therefore, the functions \(f(\theta)\) and \(g(\theta)\) are linearly independent.

Step-by-step solution

Find the first derivatives of the functions

The first thing we need to do is find the first derivative of each function with respect to \(\theta\). For \(f(\theta)\), we have: \[ f'(\theta) = \frac{d(\cos 3 \theta)}{d \theta} = -3 \sin(3 \theta) \] For \(g(\theta)\), we have: \[ g'(\theta) = \frac{d(4 \cos^3 \theta - 3 \cos \theta)}{d \theta} = -12 \cos^2 \theta \sin \theta + 3 \sin \theta \]

Calculate the Wronskian

Now, we will calculate the Wronskian \(W(f, g)\) of these two functions and their derivatives: \[ W(f, g) = \begin{vmatrix} f(\theta) & g(\theta) \\ f'(\theta) & g'(\theta) \end{vmatrix} = \begin{vmatrix} \cos 3 \theta & 4 \cos^3 \theta - 3 \cos \theta \\ -3 \sin 3 \theta & -12 \cos^2 \theta \sin \theta + 3 \sin \theta \end{vmatrix} \] To find the determinant, use the formula: \[ W(f, g) = f(\theta)g'(\theta) - g(\theta)f'(\theta) \] Substitute our functions and their derivatives: \[ W(f, g) = (\cos 3 \theta)(-12 \cos^2 \theta \sin \theta + 3 \sin \theta) - (4 \cos^3 \theta - 3 \cos \theta)(-3 \sin 3 \theta) \]

Simplify the Wronskian

Now, we will simplify the Wronskian: \begin{align*} W(f, g) &= -12 \cos^2 \theta \sin \theta \cos 3 \theta + 3 \sin \theta \cos 3 \theta + 12 \sin 3 \theta \cos^3 \theta - 9 \sin 3 \theta \cos \theta \\ &= 3 \sin \theta \cos 3 \theta - 12 \cos^2 \theta \sin \theta \cos 3 \theta + 12 \sin 3 \theta \cos^3 \theta - 9 \sin 3 \theta \cos \theta \\ \end{align*}

Determine Linear Independence or Dependence

To determine if the functions are linearly independent or dependent, we will look for any specific value of \(\theta\) that results in the Wronskian being nonzero. Let us take \(\theta = 0\). \[ W(0, 0) = 3 \sin 0 \cos 0 - 12 \cos^2 0 \sin 0 \cos 0 + 12 \sin 0 \cos^3 0 - 9 \sin 0 \cos 0 = 0 \] Since the Wronskian is zero at \(\theta = 0\), there may be a chance that the functions are linearly dependent. However, this is not enough to confirm their linear dependence, as more values of \(\theta\) need to be checked. Suppose, we take \(\theta = \frac{\pi}{12}\). \[ W\left(\frac{\pi}{12}, \frac{\pi}{12}\right) \neq 0 \] In this case, we obtain a nonzero value of the Wronskian. As the Wronskian of the functions is nonzero at some point, it implies that the functions \(f(\theta) = \cos 3 \theta\) and \(g(\theta) = 4 \cos^3 \theta - 3 \cos \theta\) are linearly independent.

Key concepts

Wronskian

The Wronskian is a critical tool used in the study of differential equations to determine the linear independence of a set of functions. Named after the Polish mathematician Józef Hoene-Wroński, it is defined as the determinant of a special matrix composed of the functions and their derivatives up to the \(n-1\)th order, where \(n\) is the number of functions being tested.

The step-by-step solution illustrates this concept by calculating the Wronskian for \(f(\theta)=\cos 3 \theta\) and \(g(\theta)=4 \cos ^{3} \theta-3 \cos \theta\). When functions are considered linearly independent, it means they cannot be expressed as a linear combination of each other. If the Wronskian is nonzero for some values of \(\theta\), this implies linear independence between \(f\) and \(g\). Conversely, if it's zero for all \(\theta\), the functions are linearly dependent.

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. They appear across various fields of science and engineering whenever a deterministic relationship involving some continuously varying quantities, and their rates of change in space and/or time, is known or postulated. This is essentially an equation that involves some function and its derivatives.

In the context of our exercise, \(f(\theta)\) and \(g(\theta)\) could represent solutions to some differential equations. Differential equations are central to understanding dynamics, oscillations, and many other natural phenomena. By studying the linear independence of solutions to a differential equation using the Wronskian, one can gain further insight into the behavior of solutions and the underlying system.

First Derivative

The first derivative of a function represents the rate at which the function's values are changing at a given point. It is a fundamental concept in calculus and analysis, symbolizing the instantaneous rate of change, or the slope of the tangent line to the function's graph at that point.

The computation of the first derivatives of \(f(\theta)\) and \(g(\theta)\) in the solution steps is a prerequisite for the calculation of the Wronskian. These derivatives serve as the building blocks for understanding more complex behavior of functions, such as curvature and optimization problems.

Linear Dependence

Linear dependence between functions or vectors refers to a relationship where one of the functions or vectors can be constructed by a linear combination of the others. If a set of functions or vectors is linearly dependent, then at least one of the members of the set can be expressed as a combination of the others.

In real-world terms, imagine having a set of instructions to get from point A to B. If one instruction can be completely described by mixing up the other instructions, it's considered dependent. Applying this to our step-by-step exercise, we initially checked if the Wronskian was zero to see if such a redundant relationship existed between \(f(\theta)\) and \(g(\theta)\). Our findings, however, concluded that they are linearly independent, which means each function provides unique information and one cannot be created by scaling and adding any multiple of the other.